Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 17:20Z
∑_{k=1}^{n} k = n(n+1)/2
∫₀^∞ e^(-z²) dz = √π / 2
479K
8β² + 2β + 0 = 0
∑_{k=1}^{n} k = n(n+1)/2
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
191K
\lim_{r\to 0} \frac{\sin r}{r} = 1
det| 11 5 ; 0 12 | = 132
1K
det| 5 9 ; 0 6 | = 30
188K
det| 7 9 ; 1 8 | = 47
416K